From 0ae85f196700e992cf0e39933d9173c661c5f634 Mon Sep 17 00:00:00 2001
From: yanglbme <szuyanglb@outlook.com>
Date: Tue, 7 Jan 2025 20:00:30 +0800
Subject: [PATCH] perf: update python solution to lc problem: No.3019

No.3019.Number of Changing Keys
---
 solution/3000-3099/3019.Number of Changing Keys/README.md    | 4 ++--
 solution/3000-3099/3019.Number of Changing Keys/README_EN.md | 4 ++--
 solution/3000-3099/3019.Number of Changing Keys/Solution.py  | 2 +-
 3 files changed, 5 insertions(+), 5 deletions(-)

diff --git a/solution/3000-3099/3019.Number of Changing Keys/README.md b/solution/3000-3099/3019.Number of Changing Keys/README.md
index 6398be4e58991..065a79f073985 100644
--- a/solution/3000-3099/3019.Number of Changing Keys/README.md	
+++ b/solution/3000-3099/3019.Number of Changing Keys/README.md	
@@ -31,7 +31,7 @@ tags:
 <pre>
 <strong>输入：</strong>s = "aAbBcC"
 <strong>输出：</strong>2
-<strong>解释：</strong> 
+<strong>解释：</strong>
 从 s[0] = 'a' 到 s[1] = 'A'，不存在按键变更，因为不计入 caps lock 或 shift 。
 从 s[1] = 'A' 到 s[2] = 'b'，按键变更。
 从 s[2] = 'b' 到 s[3] = 'B'，不存在按键变更，因为不计入 caps lock 或 shift 。
@@ -75,7 +75,7 @@ tags:
 ```python
 class Solution:
     def countKeyChanges(self, s: str) -> int:
-        return sum(a.lower() != b.lower() for a, b in pairwise(s))
+        return sum(a != b for a, b in pairwise(s.lower()))
 ```
 
 #### Java
diff --git a/solution/3000-3099/3019.Number of Changing Keys/README_EN.md b/solution/3000-3099/3019.Number of Changing Keys/README_EN.md
index ca7144d2a990f..40d4ccdb81352 100644
--- a/solution/3000-3099/3019.Number of Changing Keys/README_EN.md	
+++ b/solution/3000-3099/3019.Number of Changing Keys/README_EN.md	
@@ -30,7 +30,7 @@ tags:
 <pre>
 <strong>Input:</strong> s = &quot;aAbBcC&quot;
 <strong>Output:</strong> 2
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 From s[0] = &#39;a&#39; to s[1] = &#39;A&#39;, there is no change of key as caps lock or shift is not counted.
 From s[1] = &#39;A&#39; to s[2] = &#39;b&#39;, there is a change of key.
 From s[2] = &#39;b&#39; to s[3] = &#39;B&#39;, there is no change of key as caps lock or shift is not counted.
@@ -74,7 +74,7 @@ The time complexity is $O(n)$, where $n$ is the length of the string $s$. The sp
 ```python
 class Solution:
     def countKeyChanges(self, s: str) -> int:
-        return sum(a.lower() != b.lower() for a, b in pairwise(s))
+        return sum(a != b for a, b in pairwise(s.lower()))
 ```
 
 #### Java
diff --git a/solution/3000-3099/3019.Number of Changing Keys/Solution.py b/solution/3000-3099/3019.Number of Changing Keys/Solution.py
index bce74393a5ed6..61b3f0d8b0499 100644
--- a/solution/3000-3099/3019.Number of Changing Keys/Solution.py	
+++ b/solution/3000-3099/3019.Number of Changing Keys/Solution.py	
@@ -1,3 +1,3 @@
 class Solution:
     def countKeyChanges(self, s: str) -> int:
-        return sum(a.lower() != b.lower() for a, b in pairwise(s))
+        return sum(a != b for a, b in pairwise(s.lower()))